Describe and show your calculations on how you would prepare 1.00L of a 0.50M solution of each of the following
a.) HCl from "concentrated" (12M) reagent.
b.) Sodium Carbonate from the pure solid
PLEASE SO LOST(explanation of why you did of what woulp help so much) Thank You
Well, preparing solutions can be a bit daunting, but don't worry, I'll try to make it as fun as possible! Let's start with the first question:
a.) To prepare 1.00L of a 0.50M HCl solution from a "concentrated" (12M) reagent, we need to use the formula:
M1V1 = M2V2
M1 is the concentration of the stock solution (12M), V1 is the volume of the stock solution we need to dilute, M2 is the desired concentration (0.50M), and V2 is the final volume of the diluted solution (1.00L).
Now, let's substitute the values into the formula and solve for V1:
(12M)(V1) = (0.50M)(1.00L)
V1 = (0.50M)(1.00L) / 12M
V1 ≈ 0.042L
So, to prepare 1.00L of a 0.50M HCl solution from the concentrated reagent, you would need to measure 42 mL (0.042L) of the concentrated HCl solution and then dilute it with water to reach a final volume of 1.00L.
b.) Now, let's move on to preparing 1.00L of a 0.50M Sodium Carbonate solution from the solid compound. Sodium Carbonate has the chemical formula Na2CO3.
To calculate the mass of the solid compound needed, we can use the formula:
M = n / V
M is the molar mass of Sodium Carbonate, n is the number of moles, and V is the volume of the solution in liters.
Since we want a 0.50M solution, we need to find the number of moles:
n = M * V
n = (0.50mol/L) * 1.00L
n = 0.50mol
Next, let's find the molar mass of the compound:
M = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
M = (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
M ≈ 105.99 g/mol
Finally, let's plug in the values and calculate the mass of Sodium Carbonate needed:
mass = n * M
mass = 0.50mol * 105.99 g/mol
mass ≈ 53.00g
So, to prepare 1.00L of a 0.50M Sodium Carbonate solution, you would need to measure approximately 53.00g of the solid Sodium Carbonate and dissolve it in enough water to reach a final volume of 1.00L.
I hope this explanation brings a smile to your face, even if the calculations can be a bit overwhelming!
Sure! I will explain the step-by-step process on how to prepare 1.00L of a 0.50M solution for both cases.
a.) HCl from "concentrated" (12M) reagent:
To prepare 1.00L of a 0.50M solution of HCl, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
Step 1: Calculate the moles of HCl required:
Given that the desired molarity is 0.50M and the volume of the solution is 1.00L, we can use the rearranged formula:
moles of solute (HCl) = Molarity (M) x volume of solution (L)
moles of HCl = 0.50M x 1.00L = 0.50 moles
Step 2: Calculate the volume of concentrated HCl required:
Given that the concentrated HCl reagent has a concentration of 12M, we can use the formula:
Molarity (M) = moles of solute / volume of solution (L)
Rearranging the formula:
volume of solution (L) = moles of solute / Molarity (M)
volume of concentrated HCl = 0.50 moles / 12M = 0.0417 L = 41.7 mL
Therefore, to prepare 1.00L of a 0.50M HCl solution, you would need to measure 41.7 mL of concentrated HCl and then add water to make a total volume of 1.00L.
b.) Sodium Carbonate from the pure solid:
To prepare 1.00L of a 0.50M solution of sodium carbonate, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
Step 1: Calculate the moles of sodium carbonate required:
Given that the desired molarity is 0.50M and the volume of the solution is 1.00L, we can use the rearranged formula:
moles of solute (sodium carbonate) = Molarity (M) x volume of solution (L)
moles of sodium carbonate = 0.50M x 1.00L = 0.50 moles
Step 2: Calculate the mass of sodium carbonate required:
The molar mass of sodium carbonate (Na2CO3) is 105.99 g/mol.
mass of sodium carbonate = moles of sodium carbonate x molar mass of sodium carbonate
mass of sodium carbonate = 0.50 moles x 105.99 g/mol = 52.995 g (approximately 53 g)
Therefore, to prepare 1.00L of a 0.50M solution of sodium carbonate, you would need to measure approximately 53 grams of pure solid sodium carbonate and then add water to make a total volume of 1.00L.
Sure! I'd be happy to help explain the process of preparing these solutions. Let's start with part a.)
a.) To prepare 1.00L of a 0.50M HCl solution from a concentrated (12M) reagent, we need to calculate the volume of the concentrated solution required and then dilute it to the desired concentration.
To find the volume of 12M HCl required, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case:
C1 = 12M
V1 = ?
C2 = 0.50M
V2 = 1.00L
Let's substitute the values into the formula and solve for V1:
12M * V1 = 0.50M * 1.00L
V1 = (0.50M * 1.00L) / 12M
V1 = 0.0417L or 41.7mL
So, to prepare 1.00L of a 0.50M HCl solution from a concentrated (12M) reagent, you need to measure 41.7mL of the concentrated HCl solution and then dilute it with distilled water to a final volume of 1.00L.
Now, let's move on to part b.)
b.) To prepare 1.00L of a 0.50M Sodium Carbonate (Na2CO3) solution from the pure solid, we need to calculate the mass of Na2CO3 required and then dissolve it in water to make the solution.
The molar mass of Na2CO3 is:
Na = 23 grams/mol
C = 12 grams/mol
O = 16 grams/mol (x3)
Na2CO3 = (23x2) + 12 + (16x3) = 46 + 12 + 48 = 106 grams/mol
To find the mass of Na2CO3 required, we can use the formula:
M = (m/M) * V
Where:
M = molar concentration (0.50M)
m = mass
M = molar mass (106 g/mol)
V = volume (1.00L)
Let's substitute the values into the formula and solve for m:
0.50M = (m / 106g/mol) * 1.00L
m = 0.50M * 106g/mol
m = 53 grams
So, to prepare 1.00L of a 0.50M Sodium Carbonate solution from the pure solid, you need to measure 53 grams of Na2CO3 and then dissolve it in distilled water to a final volume of 1.00L.
I hope this explanation clarifies the process of preparing these solutions. If you have any further questions, feel free to ask!
a)
since the concentration is reduced by a factor 24, use 1/24 L of concentrated acid and add 23/24 L of water.
b)
1L of .5M contains .5 moles of Na2CO3. So, put 53g Na2CO3 in a beaker and fill with water to 1.00L of volume